Discovering Geometry Chapter 4 Test Review
HGSH
A
A
1
2
4
S
3
1 and 2 are Vertical Angles.
3 and 4 are Alternate Interior Angles
and are equal by the Parallel Line Conjecture.
Discovering Geometry Chapter 4 Test Review
HGSH
BC = (8X +3)
YZ = (7X +5)
Because of the way the triangle congruence is written
A and X are corresponding angles
C and Z are corresponding angles
BC and YZ are corresponding, So:
BC = YZ
8x + 3 = 7x + 5 , subtract 7x
x+3 = 5
x=2
, subtract 3
Discovering Geometry Chapter 4 Test Review
HGSH
Triangle Sum Conjecture
B = 180° – (60° + 45°)
B = 75°
45°
120°
120° is the exterior angle of the triangle.
By the Triangle Exterior Angle Conjecture,
120° = 90° + a
120° – 90° = a
30° = a
OR, 120° and b are supplementary
So, b= 180° - 120°
b= 60°
And, a = 180 – (90 +b) = 180 – (90-60)
a = 30°
Discovering Geometry Chapter 4 Test Review
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50°
65°
By the Triangle Exterior Angle Conjecture
x= 50° + 65°
x= 115°
Another way, by the Triangle Sum Conjecture,
the sum of the interior angles of a triangle equals 180°.
So the third angle is:
180° – ( 65° +50°) = 65°
and
x + 65° = 180° since they are supplementary
x= 180° – 65°
x= 115°
Discovering Geometry Chapter 4 Test Review
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By the Triangle Inequality Conjecture,
the sum of two of the legs has to be greater
than the third.
22cm + 29cm = 51cm and it’s not greater than 52cm.
One of the two, 22cm or 29cm legs need to increase
so that the sum of the two is greater than 52 cm
52
22
29
Side-Angle Inequality Conjecture
180° – ( 65° + 55°) = 60°
From least to greatest:
j, k, l
60°
55°
65°
Discovering Geometry Chapter 4 Test Review
HGSH
C
F
25°
25°
z
A
13°
11 cm
B
D
13°
11 cm
E
Remember that CPCTC
A  D
180° = A + 25° + 13°
180° – 38° = A
142° = A (CAB) so, by CPCTC
142° = D
By CPCTC,
AB  DE
BC  EF
AC  DF
Discovering Geometry Chapter 4 Test Review
HGSH
55°
45°
110°
B
C
88°
180°
By the Triangle Exterior Angle Conjecture
x + x= 110°
2x= 110°
x=55°
E
D
W
T
U
Discovering Geometry Chapter 4 Test Review
V
HGSH
AD  CD
 ADB   CDB
and BD is common to both triangles.
Therefore we have SAS.
Discovering Geometry Chapter 4 Test Review
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4 cm, 15 cm, 20 cm
12 cm, 11 cm, 20 cm
5 cm, 5 cm, 10 cm
14 cm, 5 cm, 20 cm
x=79°; y=101°
x=22°; y=101°
x=79°; y=68°
x=22°; y=79°
101°
Because the Triangle is an Isosceles Triangle, the base
angles are equal.
So, then y = 180° – 101° (Supplementary Angles)
y=79° Therefore,
2y+x= 180°
2(79°) + x = 180°
x = 180° – 158° = 22°
Discovering Geometry Chapter 4 Test Review
HGSH
Discovering Geometry Chapter 4 Test Review
HGSH
Discovering Geometry Chapter 4 Test Review
HGSH
100°
By the Isosceles Triangle Conjecture,
the base angles of this Isosceles triangle
are equal.
So 2x + 100° = 180°
2x = 80°
x=40°
28
By the Converse of the Isosceles Triangle Conjecture,
The Triangle is an Isosceles Triangle so the legs
that include vertex P are equal, therefore, PQ is 28 mm.
Discovering Geometry Chapter 4 Test Review
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20 ft.
By the Converse of the Isosceles Triangle Conjecture
The Triangle is an Isosceles Triangle so the legs
that include vertex G are equal, therefore,
The perimeter = 20 + 6 + 6 = 32 ft.
mC = 180- (62+58) = 60
Segment BC
62°
58°
Discovering Geometry Chapter 4 Test Review
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Segment BD, the Angle Bisector of Isosceles Triangle ABC,
is also the perpendicular bisector of segment AC making
point D the midpoint. Since segment BD starts at vertex B
and goes through midpoint D, then by definition, segment
BD it is the median of triangle ABC.
Discovering Geometry Chapter 4 Test Review
HGSH
By the Converse of the Isosceles Triangle Conjecture,
the Triangle is an Isosceles, but since the base angles
are 60°, then by the Triangle Sum Conjecture,
60 + 60 + x = 180,
So x = 60, therefore making the triangle an equiangular,
equilateral triangle. Therefore,
3y = 33 yards.
y = 11 yds.
16 in
15 in
17 in
Discovering Geometry Chapter 4 Test Review
HGSH
Discovering Geometry Chapter 4 Test Review
HGSH
Discovering Geometry Chapter 4 Test Review
HGSH
Discovering Geometry Chapter 4 Test Review
HGSH